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$E_1$ and $E_2$ are events in a probability space satisfying the following constraints:

  • $\operatorname{Pr}(E_1)=\operatorname{Pr}(E_2)$
  • $\operatorname{Pr}(E_1\cup E_2)=1$
  • $E_1$ and $E_2$ are independent.

The probability of $E_1$ is ...(a) $0$, (b) $1/4$, (c) $1/2$, (d) $1$

I think the answer should be (d) 1

Reasoning.

  1. E1 and E2 are equally likely.
  2. Sum of their probability is 1. This is possible if and only if both of their probabilities are either 1(edit: if the events are independent) or 0.5( edit: if these are dependent and exhaustive events)
  3. E1, E2 are independent events. This implies that they both doesn't belong to same same space.

Hence E1 and E2 are certain events with probability 1.

Please correct me if my reasoning or answer is wrong.

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  • $\begingroup$ See math notation guide. I replaced the image with text and formulas. $\endgroup$ – user147263 Jan 10 '15 at 5:36
  • $\begingroup$ @Fundamental: Wow!!! Thanks. I will try to use these symbols from next time. :) $\endgroup$ – Prabhakar Jan 10 '15 at 11:03
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Your reasoning for (2) is in error. Imagine a 3-sided die. The probability of $A=\{1,2\}$ occurring and $B=\{2,3\}$ occurring is equal and their union is a certain event.

You can get the answer quickly by elimination:

$$1=P(E_1 \cup E_2) < P(E_1)+P(E_2) = 2P(E_1) = 2P(E_2)$$

The strict less-than comes from being independent. So $P(E_1),P(E_2)$ are both larger than $1/2$

You can explicitly prove this, too

$$1=P(E_1 \cup E_2) = P(E_1)+P(E_2)-P(E_1)P(E_2) = 2P(E_1) - P(E_1)^2$$

$$\iff P(E_1)^2 - 2P(E_1)+1=(P(E_1)-1)^2=0$$

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  • $\begingroup$ in the example of a 3 sided die, the occurrences of any 2 events is either dependent or mutually exclusive. As per the 3rd point in the image, the events must be independent. So, I don't think we can consider this example for this question. $\endgroup$ – Prabhakar Jan 10 '15 at 5:01
  • $\begingroup$ But your reasoning for $(2)$ without having considered their independence was in error. You didnt state that until already concluding it must be $1/2$ or $1$. You would have been marked off for this in a solution. $\endgroup$ – David Peterson Jan 10 '15 at 5:03
  • $\begingroup$ Thanks. I missed it. I just updated the point 2. $\endgroup$ – Prabhakar Jan 10 '15 at 5:13
  • $\begingroup$ : according to you which option is the correct answer? $\endgroup$ – Prabhakar Jan 10 '15 at 5:17
  • $\begingroup$ If $(x-1)^2=0$ then $x$ must be.... $\endgroup$ – David Peterson Jan 10 '15 at 5:18

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